Let $G$ be a finite cyclic group. Then it is a well-known fact, that one can compute its Tate-cohomology groups from the complex $$\cdots\xrightarrow{\;\tau \;-\;\text{Id} \;}M \xrightarrow{\;\;Tr_{G}\;\;} M \xrightarrow{\;\tau \;-\;\text{Id} \;} M \xrightarrow{\;\;Tr_G\;\;} M \xrightarrow{\;\tau \;-\;\text{Id} \;}\cdots $$
where $\tau$ is multiplication by some generator of $G$ and $Tr_G(x)= \sum_{g \in G} gx$.

Can you tell me why this defines an acyclic resolution of our module? I do not see why this complex necessarily has to compute the cohomology groups.

1 Answer
1

It doesn't define an acyclic resolution (at least, unless $M$ is acyclic). But one can show that the cohomology groups that arise from that complex are isomorphic to the usual ones as follows:

The following is from Chapter VIII, $\S$4 of Serre's Local Fields (I changed the notation a bit to fit yours though):

Define a cochain complex $K$ as follows: $K^i=\mathbb{Z}[G]$ for all $i$, $d:K^i\to K^{i+1}$ is $(\tau - \text{Id})$ if $i$ is even and $Tr_G$ if $i$ is odd. For each $G$-module $A$, put $K(A)=K\otimes_{\mathbb{Z}[G]}A$. Then $K^i(A)=A$ for all $i$, with the induced maps being the same. An exact sequence $0\to A\to B\to C\to 0$ gives rise to an exact sequence of complexes $0\to K(A)\to K(B)\to K(C)\to 0$ whence to an exact cohomology sequence, and, in particular, to a coboundary operator $\delta$.

Proposition 6. The cohomological functor $\{H^q(K(-)),\delta\}$ is isomorphic to the functor $\{H^q(G,-),\delta\}$.

First of all it is clear that $\widehat{H}{}^0(G,A)=H^0(K(A))$, $\widehat{H}{}^{-1}(G,A)=H^{-1}(K(A))$, and that the coboundary operator $\delta$ relating $H^0$ to $H^{-1}$ is the same. Hence
$$H^q(K(A))=0$$
for $q=0,-1$ when $A$ is relatively projective, thence for all $q$ (as the $H^q(K(A))$ depend only on the parity of $q$). That suffices to give the isomorphism.

The last bit is using that a $G$-module is relatively projective if and only if it is relatively injective when $G$ is finite, and an abstract characterization of derived functors (of which group cohomology is an example) - see the last sentence of the section here.